And from looking at result in the question and expanding the squared term on top this suggests that ##(\textbf{p}_C \cdot \textbf{p}_D) = m_C m_D ## and I cannot see how!

I tried doing the dot product, but the only way I can see how that works is if both ##C## and ##D## have zero three-velocity, and I cant see how that could be from the question if so, and if not, I cant see how it comes about.

Also remember that the product of the two four-vectors is invariant, so you can calculate it in the center-of-mass frame instead of the lab frame.

Ok, thank you. Yeah I knew it was invariant but maybe did not fully understand the principle or implications of it. I assumed that the question set's it up so that it is from particle ##B## 's rest frame, and that is how I got ##E_a m_B## from the inner product of ##\textbf{p}_A## and ##\textbf{p}_B## since ##B## has no 3-velocity . And I assumed if I computed that bit in B's rest frame then the other bits of the equation had too to? Or is it that because 4-momentum is invariant I can calculate that one bits from B's rest frame, and then compute the other bits from the CoM frame of ##C## and ##D##?

What is the relative velocity of C and D at the minimal energy for this reaction?

Something went wrong with the formatting (missing line breaks?), but what you got at the end is correct, it just needs a last simplification step.

Thanks for replying, the formatting looks fine for me? All the line breaks are coded and are working properly, well, according to my browser at least :) . The relative velocity would be zero wouldn’t it? I.e. moving at same velocity.

Staff: Mentor

After reloading, the line breaks are now shown to me as well. Probably a weird one-time error.

The relative velocity is zero, right, this should give you the product of the two momentum vectors, especially if you use vela's hint. If something is invariant under Lorentz transformations, it does not matter in which reference frame you evaluate it - that is exactly what the invariance tells you.